You can come here to find information about what we are learning in class. I will add more information as we move forward through the curriculum. I will post up review work sheets, notes on topics we discuss in class, and links to websites that you can go to for more information or to play games. One note that I want to make is that for the problems I put on this wiki *=multiply, /=divide, and ^=exponents.
PEMDAS
PEMDAS is the acronym we use for the order of operations in math. P stands for parenthesis, E for exponents, M for multiply, D for divide, A for add, and S for subtract. We use this for complex equations such as 5^2+3(9-3)-6/2. In this equation we first do the exponents (5^2) so now we have 25+3(9-3)-6/2. Next we do the parenthesis (9-3) so we have 25+3*6-6/2. Then we can multiply and divide (3*6 and 6/2) so we have 25+18-3. The last thing we do is the addition and subtraction 25+18-3=40. Here is a website you can go to that will show you a few more examples and explain even more: PEMDAS
Now try to do these problems and we will go over them in class:
Variables are letters that we use to represent a specific number. For example in the problem x+5=19. In order to solve for x we would subtract 5 from both sides and get x=14. If you have a subtraction problem such as x-20=5 you would add 20 to both sides to get x=25. For a multiplication problem such as 5x=35, you would divide 5 from both sides to get x=7. For a division problem such as x/6=9, you would multiply by 6 on both sides and get x=54. If you have exponents in the problem you take the root of both sides to get what x equals. For example x^2=81, when you take the square root of both side you get x=9. Essentially you do the oposite operation in order to get rid of the number on the same side as the variable. As we advance in this topic we will get more complex equations to solve for the variables. The most important thing to remember is that in order to solve for the variable you want to get it alone on one side of the equal sign. Here is a website you can go to that will explain variables a little more: Variables
Now solve for x in these problems and we will go over them in class:
We will now apply what we originally learned about variables to solving for a variable in a complex equation. We will look at x^2+4(3+5)+12/4=60. When we look at this problem the first thing we will do is solve the equations that do not involve the variable in order to make it easier to solve for x. Remembering PEMDAS we will first solve the parenthesis and get x^2+4*8+12/4=60. We can't do the exponents because x is still unknown so we skip that and move onto multiply and divide. So now we have x^2+32+3=60. Next we add to get x^2+35=60. Now we add 35 to both sides to get x alone. We now have x^2=25. Finally we need to take the square root of both sides to get x=5.
Now solve for x in these problems and we will go over them in class:
Welcome to Our Math Class Wiki!
About this Wiki
You can come here to find information about what we are learning in class. I will add more information as we move forward through the curriculum. I will post up review work sheets, notes on topics we discuss in class, and links to websites that you can go to for more information or to play games. One note that I want to make is that for the problems I put on this wiki *=multiply, /=divide, and ^=exponents.PEMDAS
PEMDAS is the acronym we use for the order of operations in math. P stands for parenthesis, E for exponents, M for multiply, D for divide, A for add, and S for subtract. We use this for complex equations such as 5^2+3(9-3)-6/2. In this equation we first do the exponents (5^2) so now we have 25+3(9-3)-6/2. Next we do the parenthesis (9-3) so we have 25+3*6-6/2. Then we can multiply and divide (3*6 and 6/2) so we have 25+18-3. The last thing we do is the addition and subtraction 25+18-3=40. Here is a website you can go to that will show you a few more examples and explain even more: PEMDASNow try to do these problems and we will go over them in class:
1. 4*7-22/11+(10-4)^2 2. 6/3+5(6-3)-4^2 3. 10-8/4+3^3-3(12-7)
Variables
Variables are letters that we use to represent a specific number. For example in the problem x+5=19. In order to solve for x we would subtract 5 from both sides and get x=14. If you have a subtraction problem such as x-20=5 you would add 20 to both sides to get x=25. For a multiplication problem such as 5x=35, you would divide 5 from both sides to get x=7. For a division problem such as x/6=9, you would multiply by 6 on both sides and get x=54. If you have exponents in the problem you take the root of both sides to get what x equals. For example x^2=81, when you take the square root of both side you get x=9. Essentially you do the oposite operation in order to get rid of the number on the same side as the variable. As we advance in this topic we will get more complex equations to solve for the variables. The most important thing to remember is that in order to solve for the variable you want to get it alone on one side of the equal sign. Here is a website you can go to that will explain variables a little more: VariablesNow solve for x in these problems and we will go over them in class:
1. x-15=30 2. x+16=40 3. 12x=144 4. x/10=750 5. x^2=64
Variables in complex equations
We will now apply what we originally learned about variables to solving for a variable in a complex equation. We will look at x^2+4(3+5)+12/4=60. When we look at this problem the first thing we will do is solve the equations that do not involve the variable in order to make it easier to solve for x. Remembering PEMDAS we will first solve the parenthesis and get x^2+4*8+12/4=60. We can't do the exponents because x is still unknown so we skip that and move onto multiply and divide. So now we have x^2+32+3=60. Next we add to get x^2+35=60. Now we add 35 to both sides to get x alone. We now have x^2=25. Finally we need to take the square root of both sides to get x=5.Now solve for x in these problems and we will go over them in class:
1. 5(x-4)+33=14 2. 5^3-8(3-2)-x=30 3. x+5^2-7*2=13(6-4)